i seem to be able to prove that, the existence of a ring homomorphism $\phi:A \to B$ implies the characteristic of $B$ divides that of $A$, but Gallian says this is implied when the ring homomorphism is surjective
If you mean that I should've added all fours in the min\max functions, it is a waste of symbols, all $s$ are less than all $t$. On the wording. Should I have said "The restriction of $f$ to $[s,t]$ has its max. min..."?
Writing $R[S]$ for some ring $R$ and a random subset $S$ is nonsense notation, unless $R$ is already understood to be embedded in some larger ring $R'$, and $S$ is a subset of $R'$.
It is not incorrect, noball. I already told you my preference. This is why I don’t want to keep doing this. A comment or suggestion turns into a time-wasting battle.
if you don't have anything interesting to add, and if you don't recognize perfectly valid mathematical investigation for a learner, you have nothing of use or interest to add
this is a room associated with Math.SE officially, you're not free to make deliberate use of your moderation powers when you can't have the humility to accept you're wrong
A gentle notice: I am not engaging in any moderation activity right now. The reasons for this have been posted on the main meta, and on the math meta. However, this does not mean that you have free reign to behave badly in chat---when the strike ends, if I still have a diamond, it is likely that moderation action will be taken. Also note that @TedShifrin is a room owner here and, unless he, too, is striking, has the power to kick users from this room, or to put the room in timeout.
It's very funny, the perspective makes me think that the above picture should embed in a torus. But of course that's not true, because the Borromean link is a hyperbolic link -- these are never torus links.
I think the point is any two $(1, 1)$-curves on the standard torus in $\Bbb R^3$ actually link once in $\Bbb R^3$, i.e. they make a Hopf link
I should learn how draw better pictures digitally someday. I have a cool, pictorial solution to a problem I'd like to send to AMM, but it'll take effort to render those pictures.
It easy to make a torus with the small circles replaced by ellipses. But there isn't a sensible way to keep the small circles as circles while turning the big circles into ellipsea.
so Maschke's theorem says that if the characteristic of a field $k$ does not divide $|G|$ then reducible module and decomposable module are the same in $kG$-module?
yes, the notation is wrong. just write what you want to write as a set. R = {values of polynomials with coefficients in Z/nZ at 1/a}. this is sensible iff a and n are coprime (so 6 and 2 is actually nonsensical, not even notation-wise). and in that case, R is indeed a ring and what that ring is is just Z/nZ
@TedShifrin Thanks for your answer. I agree that I haven't talked about geodesic curvature there. But wasn't this the normal curvature I described there?
you're throwing words out with a near nil understanding of them. localization of a ring R at an element x doesn't say the only denominators are x^n. you allow fractions of the form a * b/(x^n * b), because rules of arithmetic has to be consistent for it to be a well-defined ring. which is exactly you cannot localize at zero divisors
well, you can, you just get the zero ring.
the definition of localization for zero divisors is slightly different to adjust for this
it certainly can get more confusing if there's no ring playing the role of Q here, i.e., there's no known ring that all of our constructions are playing out inside of
alright: we draw the grid, the top row has numbers 1 through 5, the subsequent rows divide by powers of two. hypothetically the grid goes on to infinity. now: suppose multiplication and addition are defined as usual but all results must be modded by 6, and from division the only thing we have is (1/n)n = 1
how many elements can we eliminate from the grid, because we discover they're equivalent to other elements of the grid?
we don't know from the start, that's the mathematical problem. can you come up with a ring structure knowing only n(1/n) = 1, and the only certitude being that you can mod n and not 1/n?
that's not what I meant, i meant that the problem begins with the fact we don't know what modding fractions means, but we know we have a/b = (a mod 6)/b
maybe you should be less confrontational before you're wrong than being confrontational and apologizing afterwards for being wrong.
people tend to dislike you otherwise. math is, after all, a social activity - not a personal ivory tower of self-righteousness where you shower your disdain towards others from